The sum of the squares of two consecutive positive numbers is 61. What is the smaller number?

2 Answers
jags
Jun 28, 2016

Smallest no is #5#. Other no is 6.

Explanation:

Let the smaller no be #n#. So the other no is #n+1#

Given: #n^2+(n+1)^2 = 61 =>#

#n^2 + n^2 + 2n + 1 - 61 = 0#

#n^2 + n^2 + 2n + 1 - 61 = 0#

#2n^2 + 2n - 60 = 0#

#n^2 + n - 30 = 0#

# (n+6)(n-5) = 0# => # #n = -6# or #n=5#

Discard #-6 # as n is positive. So, #n = 5#

Shwetank Mauria
Jun 28, 2016

The smaller number is #5#.

Explanation:

Let the smaller number be #x# and then next consecutive number is #x+1#. As the sum of the squares of these two consecutive positive numbers is #61#, we have

#x^2+(x+1)^2=61# or

#x^2+x^2+2x+1=61# or

#2x^2+2x-60=0# or (diving each side by #2#)

#x^2+x-30=0# or

#x^2+6x-5x-30=0# or

#x(x+6)-5(x+6)=0# or

#(x+6)(x-5)=0#

Hence #x=-6# or #x=5#.

But as we need positive numbers only, #x=5#.

Related questions
Impact of this question
4650 views around the world
You can reuse this answer
Creative Commons License